Brian Reed - cp's OEIS frontend

User: Brian Reed

Brian Reed has authored 2 sequences.

A198193 Replace 2^k in the binary representation of n with n+(k-L) where L = floor(log(n)/log(2)).

0, 1, 2, 5, 4, 8, 11, 18, 8, 15, 18, 28, 23, 35, 39, 54, 16, 30, 33, 50, 38, 57, 61, 83, 47, 70, 74, 100, 81, 109, 114, 145, 32, 61, 64, 96, 69, 103, 107, 144, 78, 116, 120, 161, 127, 170, 175, 221, 95, 141, 145, 194, 152, 203, 208, 262, 165, 220, 225, 283

Offset: 0

Brian Reed, Oct 26 2011

That is, if n = 2^a + 2^b + 2^c + ... then a(n) = (n+(a-L)) + (n+(b-L)) + (n+(c-L)) + ...).

			a(4) = (4+(2-2)) = 4 because int(log(4)/log(2)) = 2 and 4 = 2^2.
a(6) = (6+(2-2)) + (6+(1-2)) = 11 because int(log(6)/log(2)) = 2 and 6 = 2^2 + 2^1.

Cf. A000120, A073642, A198192.

% n is number of terms to be computed, b is the base. The examples all use b=2:
function [V] = revAddition(n,b)
   for i = 0:n
      k = i;
      if (i > 0)
         l = floor(log(i)/log(b));
      end
      s = 0;
      while(k ~= 0)
         if ((i-l) >= 0)
            s = s + mod(k,b)*(i-l);
         end
         l = l - 1;
         k = (k - mod(k,b))/b;
      end
      V(i+1) = s;
   end
end

read("transforms") :
A198193 := proc(n)
        (n-A000523(n))*wt(n)+A073642(n) ;
end proc:
seq(A198193(n),n=0..20) ; # R. J. Mathar, Nov 17 2011

Table[b = Reverse[IntegerDigits[n, 2]]; L = Length[b] - 1; Sum[b[[k]] (n + k - 1 - L), {k, Length[b]}], {n, 0, 59}] (* T. D. Noe, Nov 01 2011 *)

def A198193(n): return sum((n-i)*int(j) for i,j in enumerate(bin(n)[2:])) # Chai Wah Wu, Mar 13 2021

Let L = A000523(n), then a(n) = (n-L)*A000120(n) + A073642(n).

A198192 Replace 2^k in the binary representation of n with n-k (i.e. if n = 2^a + 2^b + 2^c + ... then a(n) = (n-a) + (n-b) + (n-c) + ...).

Original entry on oeis.org

0, 1, 1, 5, 2, 8, 9, 18, 5, 15, 16, 29, 19, 34, 36, 54, 12, 30, 31, 52, 34, 57, 59, 85, 41, 68, 70, 100, 75, 107, 110, 145, 27, 61, 62, 99, 65, 104, 106, 148, 72, 115, 117, 163, 122, 170, 173, 224, 87, 138, 140, 194, 145, 201, 204, 263, 156, 216, 219, 282, 226

Offset: 0

Brian Reed, Oct 21 2011

			a(5) = (5-2) + (5-0) = 8 because 5 = 2^2 + 2^0.
a(7) = (7-2) + (7-1) + (7-0) = 18 because 7 = 2^2 + 2^1 + 2^0.

Alois P. Heinz, Table of n, a(n) for n = 0..10000

Cf. A000120, A073642.

% n is number of terms to be computed:
function [B] = predAddition(n)
   for i = 0:n
      k = i;
      c = 0;
      s = 0;
      while(k ~= 0)
         if ((i - c) >= 0)
            s = s + mod(k,2)*(i-c);
         end
         c = c + 1;
         k = (k - mod(k,2))/2;
      end
      B(i+1) = s;
   end
end

b:= (n, k)-> `if`(n=0, 0, k*(n mod 2)+b(floor(n/2), k-1)):
a:= n-> b(n, n):
seq(a(n), n=0..100);  # Alois P. Heinz, Oct 25 2011

a(n) = n*A000120(n) - A073642(n). - Franklin T. Adams-Watters, Oct 22 2011

a(n) = b(n,n) with b(0,k) = 0, b(n,k) = k*(n mod 2) + b(floor(n/2),k-1) for n>0. - Alois P. Heinz, Oct 25 2011

cp's OEIS Frontend

User: Brian Reed

A198193 Replace 2^k in the binary representation of n with n+(k-L) where L = floor(log(n)/log(2)).

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

MATLAB

Maple

Mathematica

Python

Formula